The radius of the in circle of a triangle is 4cm and the segments into which one side is divided by the point of contact are 6cm and 8cm.  Determine the other two sides of the triangle.(Ans: 15, 13)

Ans:    a = BC = x + 8

b = AC = 6 + 8 = 14cm

c = AB = x + 6

Semi - perimeter = a + b + c/2

=  BC + AC + AB/2

= x + 8 +14 + x + 6/2

= 2 x + 28/2

= x + 14

=   √( x + 14)(48x)  .................(1)

Area of ΔABC = area ΔAOB + area BOC + area ΔAOC

area ?AOC = ( 1/2b.h) = 1/2 x 4 x 14

= 28

On substituting we get

∴ area ΔABC = area ΔAOC + area ΔBOC + area ΔAOB

= 4x + 56                     ...............(2)

From (1) and (2)

4x + 56 =

Simplify we get     x = 7

∴AB = x + 6 = 7 + 6 = 13cm

∴BC = x + 8 = 7 + 8 = 15cm